A160050 Numerator of the Harary number for the star graph s_n.

0, 1, 5, 9, 7, 10, 27, 35, 22, 27, 65, 77, 45, 52, 119, 135, 76, 85, 189, 209, 115, 126, 275, 299, 162, 175, 377, 405, 217, 232, 495, 527, 280, 297, 629, 665, 351, 370, 779, 819, 430, 451, 945, 989, 517, 540, 1127, 1175, 612, 637, 1325, 1377, 715, 742, 1539

Offset: 1

Author

Eric W. Weisstein, Apr 30 2009

Keywords

• nonn
• easy
• frac

Examples

			0, 1, 5/2, 9/2, 7, 10, 27/2, 35/2, 22, 27, ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Eric Weisstein's World of Mathematics, Harary Index.
• Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Crossrefs

Cf. A130658 (denominators), A033954 (quadrisection), A001107 (quadrisection), A181890 (quadrisection).

Cf. A000217, A060819, A064038.

Programs

Magma

m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x^2*(-1 - 2*x - 5*x^4 + 3*x^5 - 2*x^6 + x^7)/((x - 1)^3*(x^2 + 1)^3))); // G. C. Greubel, Sep 21 2018

Mathematica

f[n_] := n/GCD[n, 4]; Array[f[#] f[# + 3] &, 58]
Rest[CoefficientList[Series[x^2*(-1 - 2*x - 5*x^4 + 3*x^5 - 2*x^6 + x^7)/((x - 1)^3*(x^2 + 1)^3), {x, 0, 50}], x]] (* G. C. Greubel, Sep 21 2018 *)

PARI

s=vector(40,n,1/4*(n+2)*(n-1)) /* fractions */
vector(#s,n,numerator(s[n])) /* this sequence */ \\ Joerg Arndt, Jan 04 2011

PARI

x='x+O('x^50); concat([0], Vec(x^2*(-1 - 2*x - 5*x^4 + 3*x^5 - 2*x^6 + x^7)/((x - 1)^3*(x^2 + 1)^3))) \\ G. C. Greubel, Sep 21 2018

Formula

Numerator of (1/4)*(n+2)*(n-1). - Joerg Arndt, Jan 04 2011

It appears that a(n + 1) = A060819(n-1) * A060819(n + 2). - Paul Curtz, Dec 23 2010 [Corrected by Joerg Arndt, Jan 04 2011]

G.f.: x^2*(-1-2*x-5*x^4+3*x^5-2*x^6+x^7) / ( (x-1)^3*(x^2+1)^3 ). - R. J. Mathar, Jan 04 2011

a(1+4*n) = (A000217(n+1)-1)/2, a(2+4*n) = (A000217(n+2)-1)/2, a(3+4*n) = A000217(n+3)-1, a(4+4*n) = A000217(n+4)-1. - Paul Curtz, Dec 23 2010.

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). This is not the shortest recurrence. -Paul Curtz, Mar 27 2011

a(1+3*n) = numerator of 9*n*(n+1)/4 = 9*A064038(1+n). - Paul Curtz, Apr 06 2011

a(n) = (n-1)*(n+2)*(3-i^((n-2)*(n-1)))/8, where i=sqrt(-1). - Bruno Berselli, Apr 07 2011, corrected by Vaclav Kotesovec, Aug 09 2022

Sum_{n>=2} 1/a(n) = 13/9 + Pi/6. - Amiram Eldar, Aug 09 2022

Extensions

Edited by N. J. A. Sloane, Dec 23 2010